What is the sum of $\left(\dfrac{1}{3}\right) + \left(\dfrac{1}{3}\right)^2 + \left(\dfrac{1}{3}\right)^3 + \left(\dfrac{1}{3}\right)^4$?
Solution: This 4-term geometric series has first term $a_0 = \frac13$ and ratio $r=\frac13$, so it has value  \begin{align*}
\dfrac{\dfrac13\left(1-\left(\dfrac13\right)^{4}\right)}{1-\frac13} &= \dfrac{\dfrac13(1-\left(\dfrac13\right)^{4})}{\dfrac23}\\
&=\dfrac12\left(1-\left(\dfrac13\right)^{4}\right)\\
&=\dfrac12\left(\dfrac{80}{81}\right)\\
&=\boxed{\dfrac{40}{81}}.
\end{align*}